hdoj 2717 Catch That Cow

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

 

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

讲解：一农夫追牛，牛很笨，他就在原地等着农夫来抓他，并且这是一条直线，农夫很容易就能找到它的，然而农夫却只有三种选择，退一步，走一步，或者走到当前位置的2倍；于是乎我们可以用搜索来解决；

代码如下：

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int map[200010];
struct T
{
    int x,step;
};
int dfs(T now)
{queue< T >q;
    T end;
    q.push(now);
    while(!q.empty())
    {
        end=q.front();
          q.pop();
        map[end.x]=1;
    if(end.x==m)
     {return end.step;}
     //如果不是走一步一判断很容易超出内存的
     now.x=end.x+1;//前进一步，存入队列；
         if(end.x>=0 && end.x<=100000 && map[now.x]==0)
         {
             now.step=end.step+1;
             map[now.x]==1;
             q.push(now);
         }
     now.x=end.x-1;//后退一步
     if(end.x>=0 && end.x<=100000 && map[now.x]==0)
     {
         now.step=end.step+1;
         map[now.x]==1;
         q.push(now);
     }
     now.x=end.x*2;//前进2倍的位置
     if(end.x>=0 && end.x<=100000 && map[now.x]==0)
     {
         now.step=end.step+1;
         map[now.x]==1;
         q.push(now);
     }
    }
    return 0;
}
int main()
{
    T now;
    while(cin>>n>>m)
    {
        if(n>=m)//如果n大于m则只能后退了；
       {
           cout<<n-m<<endl;continue;
       }
       else
       {
        memset(map,0,sizeof(map));
        now.x=n;now.step=0;
        int mm=dfs(now);
        cout<<mm<<endl;
       }
    }
    return 0;
}